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I'm trying to evaluate the given limit without using L'hopitals rule. $$\lim_{x \to 1}\frac{1-x^2}{\sin(\pi x)}$$ Replacing $x$ by $1$ leads to $\frac{0}{0}$.

I have tried multiplying by $\frac{1+x^2}{1+x^2}$ and resoving the $\sin x$ factor by doing:

$$\frac{x}{\sin x} \cdot \frac{(\frac{1}{x} -x)}{\pi}$$

so $\frac{x}{\sin x} = 1$ but then the result still is $\frac{0}{\pi}$. All other options I see also lead to $\frac{0}{0}$ and I'm too sure how to proceed. Any hints are very welcome.

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    You can't factor the $\pi$ out of the $\sin$ function, it is an argument of the function! – 2017-01-19
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    Since sin is a bounded function, you may try using the Squeeze theorem. – 2017-01-19
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    You might want to precede the sine function name with a backslash, which will convert an ordinary sequence of letters, rendered in italics, hence looking as a simple multiplication of 's', 'i' and 'n' (`sin` → $sin = s\cdot i\cdot n$) into a _symbol_, rendered in upright font (`\sin` → $\sin$). – 2017-01-19

2 Answers 2

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Using the substitution $x=1-y$ and the property $\sin(\pi-z)=\sin(z)$ we have

$$ \lim_{x\to 1}\frac{(1-x)(1+x)}{\sin(\pi x)}=\lim_{y\to 0}\frac{y(2-y)}{\sin(\pi y)}=\lim_{y\to 0}\frac{\pi y}{\sin(\pi y)}\cdot\frac{2-y}{\pi}=\color{red}{\frac{2}{\pi}}.$$

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    Is it coincidence that, after applying a substitution, the new variable always goes to 0 when rewriting the limit? As $y$ does in this particular case. – 2017-01-19
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    @TheAlPaca02: it is not a coincidence, I chose the substitution in order that when $x\to 1$ it happens that $y\to 0$. – 2017-01-19
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    Ah I believe I see it now, this is found by substituting $x$ in $x=1-y$ by $1$, right? – 2017-01-19
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Write $$\frac{1-x^2}{\sin\pi x}=\frac{1-x^2}{\sin(\pi-\pi x)}=\frac{1-x^2}{\sin\pi(1- x)}=\frac{(1-x)(1+x)}{2\sin\frac{\pi(1- x)}{2}\cos\frac{\pi(1- x)}{2}}=\frac12\times\frac{2}{\pi}\frac{\frac{\pi(1- x)}{2}}{\sin\frac{\pi(1- x)}{2}}\times\frac{1+x}{\cos\frac{\pi(1- x)}{2}}$$ so $t=1-x$ we have $$\lim_{x\to1}\frac{1-x^2}{\sin\pi x}=\lim_{t\to0}\frac12\times\frac{2}{\pi}\frac{\frac{\pi t}{2}}{\sin\frac{\pi t}{2}}\times\frac{2-t}{\cos\frac{\pi t}{2}}=\frac12\times\frac{2}{\pi}\times\frac{2}{1}=\color{red}{\frac{2}{\pi}}$$